Answer
$$\frac{dr}{d\theta}=-\frac{e^{r\theta}r+\csc^2\theta}{e^{r\theta}\theta+\sin r}$$
Work Step by Step
$$\cos r+\cot\theta=e^{r\theta}$$
1) Differentiate both sides of the equation with respect to $\theta$:
$$\frac{d}{d\theta}(\cos r)+\frac{d}{d\theta}(\cot\theta)=\frac{d}{d\theta}(e^{r\theta})$$
$$-\sin r\frac{dr}{d\theta}-\csc^2\theta=e^{r\theta}\frac{d}{d\theta}(r\theta)$$
$$-\sin r\frac{dr}{d\theta}-\csc^2\theta=e^{r\theta}(r+\theta\frac{dr}{d\theta})$$
$$-\sin r\frac{dr}{d\theta}-\csc^2\theta=e^{r\theta}r+e^{r\theta}\theta\frac{dr}{d\theta}$$
2) Collect all the terms with $dr/d\theta$ onto one side and solve for $dr/d\theta$:
$$e^{r\theta}\theta\frac{dr}{d\theta}+\sin r\frac{dr}{d\theta}=-e^{r\theta}r-\csc^2\theta$$
$$\frac{dr}{d\theta}(e^{r\theta}\theta+\sin r)=-e^{r\theta}r-\csc^2\theta$$
$$\frac{dr}{d\theta}=-\frac{e^{r\theta}r+\csc^2\theta}{e^{r\theta}\theta+\sin r}$$
